
which is a Poisson process with intensity measure \(\alpha \lambda _B\) .
Now we argue exactly as in the proof of [1]}.
We take a sequence \((B_n)_n\)  of bounded and closed sets
such that \(B_n\uparrow {\mathbb {X}}\) . Then
\(\chi ^{B_n}\overset{d}{\rightarrow }\xi \) ,
\(\chi ^{\prime B_n}\overset{d}{\rightarrow }\xi ^{\prime }\) 
and \(\Phi ^{B_n}\overset{d}{\rightarrow }\eta \) ,
where \(\eta \)  is a Poisson process with intensity measure
\(\alpha \lambda \)  and where we refer to [2]}
for the theory of weak convergence of point process distributions.
It follows from [3]}
that the above sequences are all tight. A standard argument
shows that \((\chi ^{B_n},\chi ^{\prime B_n},\Phi ^{B_n})\) , \(n\in {\mathbb {N}}\) ,
is tight and hence converges in distribution to
\((\tilde{\xi },\tilde{\xi }^{\prime }, \eta )\)  along some subsequence.
Then \(\xi \overset{d}{=}\tilde{\xi }\)  and \(\xi ^{\prime }\overset{d}{=}\tilde{\xi }^{\prime }\) .
Moreover, \(\eta \)  is a Poisson process with
intensity measure \(\alpha \lambda \) . We can assume that
\((\tilde{\xi },\tilde{\xi }^{\prime }, \eta )\)  is defined on the original
probability space \((\Omega ,{\mathcal {F}},\mathbb {P})\) .
Since the set \(\lbrace (\mu ,\mu ^{\prime },\psi )\in {\mathbf {N}}^3:\mu \le \psi ,\mu ^{\prime }\le \psi \rbrace \) 
is closed, it follows from the Portmanteau theorem that
\(\mathbb {P}(\tilde{\xi }\le \eta ,\tilde{\xi }^{\prime }\le \eta )=1\) .
\(\Box \)
